chapter 11 dual nature of radiation and matter

7/21/2019 Chapter 11 Dual Nature of Radiation and Matter

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Question 11.1:

Find the

maximum frequency, and

minimum wavelength of X-rays produced by 30 kV electrons.

Answer

Potential of the electrons, V = 30 kV = 3 × 104 V

Hence, energy of the electrons, E = 3 × 104 eV

Where,

e = Charge on an electron = 1.6 × 10−19 C

(a)Maximum frequency produced by the X-rays = ν

The energy of the electrons is given by the relation:

E = hν

Where,

h = Planck’s constant = 6.626 × 10−34 Js

Hence, the maximum frequency of X-rays produced is

(b)The minimum wavelength produced by the X-rays is given as:



Hence, the minimum wavele

Question 11.2:

The work function of caesiuincident on the metal surface

maximum kinetic energy of t

Stopping potential, and

maximum speed of the emitt

Answer

Work function of caesium m

Frequency of light,

(a)The maximum kinetic ene

Where,

h = Planck’s constant = 6.62

gth of X-rays produced is 0.0414 nm.

metal is 2.14 eV. When light of frequency 6 ×, photoemission of electrons occurs. What is the

he emitted electrons,

d photoelectrons?

etal,

rgy is given by the photoelectric effect as:

× 10−34 Js

1014 Hz is



Photoelectric cut-off voltage,

The maximum kinetic energ

Where,

e = Charge on an electron =

Therefore, the maximum kinexperiment is 2.4 × 10−19 J.

Question 11.4:

Monochromatic light of wav power emitted is 9.42 mW.

Find the energy and moment

How many photons per seco(Assume the beam to have u

How fast does a hydrogen atthat of the photon?

Answer

Wavelength of the monochr

Power emitted by the laser,

V 0 = 1.5 V

of the emitted photoelectrons is given as:

.6 × 10−19 C

etic energy of the photoelectrons emitted in the

length 632.8 nm is produced by a helium-neon

m of each photon in the light beam,

d, on the average, arrive at a target irradiated biform cross-section which is less than the targe

m have to travel in order to have the same mo

matic light, λ = 632.8 nm = 632.8 × 10−9 m

= 9.42 mW = 9.42 × 10−3 W

given

laser. The

this beam?area), and

entum as



Planck’s constant, h = 6.626 × 10−34 Js

Speed of light, c = 3 × 108 m/s

Mass of a hydrogen atom, m = 1.66 × 10−27 kg

(a)The energy of each photon is given as:

The momentum of each photon is given as:

(b) Number of photons arriving per second, at a target irradiated by the beam = n

Assume that the beam has a uniform cross-section that is less than the target area.

Hence, the equation for power can be written as:

(c)Momentum of the hydrogen atom is the same as the momentum of the photon,

Momentum is given as:

Where,

v = Speed of the hydrogen atom



Therefore, every second,

Question 11.6:

In an experiment on photoelof incident light is found to b

Answer

The slope of the cut-off volta

Where,

e = Charge on an electron =

h = Planck’s constant

Therefore, the value of Planc

Question 11.7:

A 100 W sodium lamp radiatthe centre of a large sphere twavelength of the sodium ligwith the sodium light? (b) At

photons are incident per square metre o

ctric effect, the slope of the cut-off voltage verse 4.12 × 10−15 V s. Calculate the value of Planc

ge (V ) versus frequency (ν) of an incident light

.6 × 10−19 C

k’s constant is

es energy uniformly in all directions. The lampat absorbs all the sodium light which is incidenht is 589 nm. (a) What is the energy per photonwhat rate are the photons delivered to the sphe

n earth.

us frequency’s constant.

is given as:

is located att on it. Theassociatede?



Answer

Power of the sodium lamp, P

Wavelength of the emitted s

Planck’s constant, h = 6.626

Speed of light, c = 3 × 108 m

(a)The energy per photon as

(b) Number of photons delive

The equation for power can

Therefore, every second,

Question 11.8:

= 100 W

dium light, λ = 589 nm = 589 × 10−9 m

× 10−34 Js

s

ociated with the sodium light is given as:

red to the sphere = n

e written as:

photons are delivered to the sphere.



The threshold frequency for1014 Hz is incident on the me

Answer

Threshold frequency of the

Frequency of light incident o

Charge on an electron, e = 1.


Cut-off voltage for the photo

The equation for the cut-off

Therefore, the cut-off voltag

Question 11.9:

The work function for a certaemission for incident radiati

Answer

certain metal is 3.3 × 1014 Hz. If light of frequ tal, predict the cutoff voltage for the photoelect

etal,

n the metal,

6 × 10−19 C

× 10−34 Js

electric emission from the metal =

nergy is given as:

for the photoelectric emission is

in metal is 4.2 eV. Will this metal give photoeln of wavelength 330 nm?

ncy 8.2 ×ic emission.

ctric



No

Work function of the metal,



Wavelength of the incident r


The energy of the incident p

It can be observed that the eof the metal. Hence, no phot

Question 11.10:

Light of frequency 7.21 × 10maximum speed of 6.0 × 105

frequency for photoemission

Answer

Frequency of the incident ph

Maximum speed of the elect

6 × 10−19 C

× 10−34 Js

diation, λ = 330 nm = 330 × 10−9 m

s

oton is given as:

ergy of the incident radiation is less than the welectric emission will take place.

14 Hz is incident on a metal surface. Electronsm/s are ejected from the surface. What is the thof electrons?

oton,

ons, v = 6.0 × 105 m/s

rk function

ith areshold




Mass of an electron, m = 9.1

For threshold frequency ν0, t

Therefore, the threshold freq

Question 11.11:

Light of wavelength 488 nm photoelectric effect. When listopping (cut-off) potential omaterial from which the emi

Answer

Wavelength of light produce

= 488 × 10−9 m

Stopping potential of the pho

1eV = 1.6 × 10−19 J

∴ V 0 =

Planck’s constant, h = 6.6 ×

× 10−34 Js

× 10−31 kg

e relation for kinetic energy is written as:

ency for the photoemission of electrons is

is produced by an argon laser which is used in tht from this spectral line is incident on the emi

f photoelectrons is 0.38 V. Find the work functiter is made.

by the argon laser, λ = 488 nm

toelectrons, V 0 = 0.38 V

0−34 Js

heter, theon of the




Speed of light, c = 3 × 10 m/

From Einstein’s photoelectriof the material of the emitter

Therefore, the material with

Question 11.12:

Calculate the

momentum, and

de Broglie wavelength of the

Answer

Potential difference, V = 56


Mass of an electron, m = 9.1


6 × 10−19 C

effect, we have the relation involving the woras:

hich the emitter is made has the work function

electrons accelerated through a potential differ

0−34 Js

× 10−31 kg

6 × 10−19 C

function Φ0

of 2.16 eV.

nce of 56 V.



At equilibrium, the kinetic ei.e., we can write the relation

The momentum of each acce

p = mv

= 9.1 × 10−31 × 4.44 × 106

= 4.04 × 10−24 kg m s−1

Therefore, the momentum of

De Broglie wavelength of anrelation:

Therefore, the de Broglie wa

Question 11.13:

What is the

momentum,

speed, and

ergy of each electron is equal to the acceleratinfor velocity (v) of each electron as:

lerated electron is given as:

each electron is 4.04 × 10−24 kg m s−1.

electron accelerating through a potential V , is g

elength of each electron is 0.1639 nm.

g potential,

iven by the



de Broglie wavelength of an electron with kinetic energy of 120 eV.

Answer

Kinetic energy of the electron, E k = 120 eV


Mass of an electron, m = 9.1 × 10−31 kg

Charge on an electron, e = 1.6 × 10−19 C

For the electron, we can write the relation for kinetic energy as:

Where,

v = Speed of the electron

Momentum of the electron, p = mv

= 9.1 × 10−31 × 6.496 × 106

= 5.91 × 10−24 kg m s−1

Therefore, the momentum of the electron is 5.91 × 10−24 kg m s−1.

Speed of the electron, v = 6.496 × 106 m/s

De Broglie wavelength of an electron having a momentum p, is given as:




Question 11.14:

The wavelength of light frokinetic energy at which

an electron, and

a neutron, would have the sa

Answer

Wavelength of light of a sodi

Mass of an electron, me= 9.1

Mass of a neutron, mn= 1.66


For the kinetic energy K , of arelation:

We have the relation for de

elength of the electron is 0.112 nm.

the spectral emission line of sodium is 589 nm

e de Broglie wavelength.

um line, λ = 589 nm = 589 × 10−9 m

× 10−31 kg

× 10−27 kg

0−34 Js

n electron accelerating with a velocity v, we ha

roglie wavelength as:

. Find the

e the



Substituting equation (2) in e

Hence, the kinetic energy of

Using equation (3), we can


Question 11.15:

What is the de Broglie wavel

quation (1), we get the relation:

he electron is 6.9 × 10−25J or 4.31 μeV.

rite the relation for the kinetic energy of the ne

he neutron is 3.78 × 10−28 J or 2.36 neV.

ength of

tron as:



a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,

a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and

a dust particle of mass 1.0 × 10−9 kg drifting with a speed of 2.2 m/s?

Answer

(a)Mass of the bullet, m = 0.040 kg

Speed of the bullet, v = 1.0 km/s = 1000 m/s


De Broglie wavelength of the bullet is given by the relation:

Mass of the ball, m = 0.060 kg

Speed of the ball, v = 1.0 m/s

De Broglie wavelength of the ball is given by the relation:

(c)Mass of the dust particle, m = 1 × 10−9 kg

Speed of the dust particle, v = 2.2 m/s

De Broglie wavelength of the dust particle is given by the relation:



Question 11.16:An electron and a photon eac

their momenta,

the energy of the photon, an

the kinetic energy of electro

Answer

Wavelength of an electron

= 1 × 10−9 m


The momentum of an eleme

It is clear that momentum dewavelengths of an electron a

The energy of a photon is gi

h have a wavelength of 1.00 nm. Find

.

10−34 Js

tary particle is given by de Broglie relation:

ends only on the wavelength of the particle. Sid a photon are equal, both have an equal mom

en by the relation:

ce thentum.



Where,


Therefore, the energy of the

The kinetic energy ( K ) of an

Where,

m = Mass of the electron = 9.

p = 6.63 × 10−25 kg m s−1


Question 11.17:

For what kinetic energy of a10−10 m?

Also find the de Broglie wav

having an average kinetic en

Answer

s

hoton is 1.243 keV.

electron having momentum p,is given by the rel

.1 × 10−31 kg

he electron is 1.51 eV.

neutron will the associated de Broglie wavelen

elength of a neutron, in thermal equilibrium wit

rgy of (3/2) kT at 300 K.

ation:

th be 1.40 ×

h matter,



De Broglie wavelength of the neutron, λ = 1.40 × 10−10 m

Mass of a neutron, mn = 1.66 × 10−27 kg

Planck’s constant, h = 6.6 × 10−34

Js

Kinetic energy ( K ) and velocity (v) are related as:

… (1)

De Broglie wavelength ( λ) and velocity (v) are related as:

Using equation (2) in equation (1), we get:

Hence, the kinetic energy of the neutron is 6.75 × 10−21 J or 4.219 × 10−2 eV.

Temperature of the neutron, T = 300 K

Boltzmann constant, k = 1.38 × 10−23 kg m2 s−2 K −1

Average kinetic energy of the neutron:

The relation for the de Broglie wavelength is given as:




Question 11.18:

Show that the wavelength ofwavelength of its quantum (p

Answer

The momentum of a photon

Where,

λ = Wavelength of the electr

c = Speed of light


elength of the neutron is 0.146 nm.

electromagnetic radiation is equal to the de Brohoton).

aving energy (hν) is given as:

magnetic radiation

glie



De Broglie wavelength of th

Where,

m = Mass of the photon

v = Velocity of the photon

Hence, it can be inferred froelectromagnetic radiation is

Question 11.19:

What is the de Broglie wavelthe molecule is moving with(Atomic mass of nitrogen =

Answer

Temperature of the nitrogen

Atomic mass of nitrogen = 1

Hence, mass of the nitrogen

But 1 u = 1.66 × 10−27 kg

∴m = 28.0152 ×1.66 × 10−27


Boltzmann constant, k = 1.3

photon is given as:

equations (i) and (ii) that the wavelength of thqual to the de Broglie wavelength of the photo

ength of a nitrogen molecule in air at 300 K? Athe root-mean square speed of molecules at this4.0076 u)

olecule, T = 300 K

.0076 u

olecule, m = 2 × 14.0076 = 28.0152 u

g

10−34 Js

× 10−23 J K −1

e.

ssume thattemperature.



We have the expression that

with the root mean square sp

Hence, the de Broglie wavel


Question 11.20:

Estimate the speed with whitube impinge on the collectothe emitter. Ignore the smallelectron, i.e., its e/m is given

Use the same formula you eof 10 MV. Do you see what i

Answer

(a)Potential difference acros

elates mean kinetic energy of the nitro

eed as:

ngth of the nitrogen molecule is given as:

elength of the nitrogen molecule is 0.028 nm.

h electrons emitted from a heated emitter of anmaintained at a potential difference of 500 V

initial speeds of the electrons. The specific char

to be 1.76 × 1011 C kg−1.

ploy in (a) to obtain electron speed for an colles wrong? In what way is the formula to be mod

the evacuated tube, V = 500 V

en molecule

evacuatedith respect toe of the

ctor potentialfied?



Specific charge of an electron, e/m = 1.76 × 1011 C kg−1

The speed of each emitted electron is given by the relation for kinetic energy as:

Therefore, the speed of each emitted electron is

(b)Potential of the anode, V = 10 MV = 10 × 106

V

The speed of each electron is given as:

This result is wrong because nothing can move faster than light. In the above formula, theexpression (mv

2/2) for energy can only be used in the non-relativistic limit, i.e., for v <<c.

For very high speed problems, relativistic equations must be considered for solving them.In the relativistic limit, the total energy is given as:

E = mc2

Where,

m = Relativistic mass

m0 = Mass of the particle at rest



Kinetic energy is given as:

K = mc2

− m0c2

Question 11.21:

A monoenergetic electron bemagnetic field of 1.30 × 10−4

circle traced by the beam, gi

Is the formula you employ inelectron beam? If not, in wha

[Note: Exercises 11.20(b) an beyond the scope of this boo point that the formulas you u

speeds or energies. See answmeans.]

Answer

(a)Speed of an electron, v =

Magnetic field experienced b

Specific charge of an electro

Where,

e = Charge on the electron =

m = Mass of the electron = 9.

The force exerted on the elec

θ = Angle between the magn

The magnetic field is normal

am with electron speed of 5.20 × 106 m s−1 is suT normal to the beam velocity. What is the raden e/m for electron equals 1.76 × 1011 C kg−1.

(a) valid for calculating radius of the path of at way is it modified?

d 11.21(b) take you to relativistic mechanics w. They have been inserted here simply to emph

se in part (a) of the exercises are not valid at ve

ers at the end to know what ‘very high speed or

.20 × 106 m/s

y the electron, B = 1.30 × 10−4 T

, e/m = 1.76 × 1011 C kg−1

1.6 × 10−19 C

.1 × 10−31 kg−1

tron is given as:

etic field and the beam velocity

to the direction of beam.

bject to aus of the

0 MeV

ich isasise they high

energy’



The beam traces a circular path of radius, r . It is the magnetic field, due to its bending

nature, that provides the centripetal force for the beam.

Hence, equation (1) reduces to:

Therefore, the radius of the circular path is 22.7 cm.

Energy of the electron beam, E = 20 MeV

The energy of the electron is given as:

This result is incorrect because nothing can move faster than light. In the above formula,the expression (mv

2/2) for energy can only be used in the non-relativistic limit, i.e., for v

<< c

When very high speeds are concerned, the relativistic domain comes into consideration.

In the relativistic domain, mass is given as:



Where,

= Mass of the particle at

Hence, the radius of the circ

Question 11.22:

An electron gun with its coll bulb containing hydrogen ga2.83 × 10−4 T curves the path

path can be viewed because telectrons, and emitting lighttube’ method. Determine e/m

Answer

Potential of an anode, V = 10

Magnetic field experienced b

Radius of the circular orbit r

Mass of each electron = m

Charge on each electron = e

Velocity of each electron = v

The energy of each electron i

est

lar path is given as:

ctor at a potential of 100 V fires out electrons iat low pressure (∼10−2 mm of Hg). A magnetiof the electrons in a circular orbit of radius 12.

he gas ions in the path focus the beam by attracy electron capture; this method is known as thfrom the data.

0 V

y the electrons, B = 2.83 × 10−4 T

= 12.0 cm = 12.0 × 10−2

m

s equal to its kinetic energy, i.e.,

a sphericalfield ofcm. (The

ing‘fine beam



It is the magnetic field, due t

for the beam. H

Centripetal force = Magnetic

Putting the value of v in equa

Therefore, the specific charg

Question 11.23:

An X-ray tube produces a coat 0.45 Å. What is the maxi

From your answer to (a), guerequired in such a tube?

Answer

its bending nature, that provides the centripeta

nce, we can write:

force

tion (1), we get:

e ratio (e/m) is

tinuous spectrum of radiation with its short waum energy of a photon in the radiation?

ss what order of accelerating voltage (for electr

l force

velength end

ons) is



Wavelength produced by an



The maximum energy of a p

Therefore, the maximum ene

Accelerating voltage provideray of 27.6 keV, the incidentenergy. Hence, an acceleratiX-rays.

Question 11.24:

In an accelerator experimentcertain event is interpreted aBeV into two γ-rays of equal

(1BeV = 109 eV)

Answer

Total energy of two γ-rays:

E = 10. 2 BeV

= 10.2 × 109 eV

= 10.2 × 109 × 1.6 × 10−10 J

-ray tube,

× 10−34 Js

s

oton is given as:

rgy of an X-ray photon is 27.6 keV.

s energy to the electrons for producing X-rays.electrons must possess at least 27.6 keV of king voltage of the order of 30 keV is required for

on high-energy collisions of electrons with posiannihilation of an electron-positron pair of totaenergy. What is the wavelength associated wit

o get an X- tic electric

producing

trons, al energy 10.2each γ-ray?



Hence, the energy of each γ-

Planck’s constant,

Speed of light,

Energy is related to wavelen

Therefore, the wavelength as

Question 11.25:

Estimating the following twyou why radio engineers dotells you why our eye can ne

The number of photons emit power, emitting radiowaves

The number of photons enterminimum intensity of white larea of the pupil to be about6 × 1014 Hz.

Answer

Power of the medium wave t

ay:

th as:

sociated with each γ-ray is

numbers should be interesting. The first numbot need to worry much about photons! The secer ‘count photons’, even in barely detectable li

ed per second by a Medium wave transmitter of wavelength 500 m.

ing the pupil of our eye per second correspondiight that we humans can perceive (∼10−10 W m

.4 cm2, and the average frequency of white lig

ansmitter, P = 10 kW = 104 W = 104 J/s

r will tellnd numberht.

10 kW

g to the2). Take thet to be about



Hence, energy emitted by the transmitter per second, E = 104

Wavelength of the radio wave, λ = 500 m

The energy of the wave is given as:

Where,

h = Planck’s constant = 6.6 × 10−34 Js

c = Speed of light = 3 × 108 m/s

Let n be the number of photons emitted by the transmitter.

∴nE 1 = E

The energy ( E 1) of a radio photon is very less, but the number of photons (n) emitted persecond in a radio wave is very large.

The existence of a minimum quantum of energy can be ignored and the total energy of aradio wave can be treated as being continuous.

Intensity of light perceived by the human eye, I = 10−10 W m−2

Area of a pupil, A = 0.4 cm

2

= 0.4 × 10

−4

m

2

Frequency of white light, ν= 6 × 1014 Hz

The energy emitted by a photon is given as:

E = h ν

Where,



h = Planck’s constant = 6.6 ×

∴ E = 6.6 × 10−34 × 6 × 1014

= 3.96 × 10−19 J

Let n be the total number of

The total energy per unit for

E = n × 3.96 × 10−19 J s−1 m−

The energy per unit area per

∴ E = I

n × 3.96 × 10−19 = 10−10

= 2.52 × 108 m2 s−1

The total number of photons

n A = n × A

= 2.52 × 108 × 0.4 × 10−4

= 1.008 × 104 s−1

This number is not as large ahuman eye to never see the i

Question 11.26:

Ultraviolet light of wavelengcell made of molybdenum mfunction of the metal. Howred light of wavelength 6328

Answer

10−34 Js

hotons falling per second, per unit area of the p

n falling photons is given as:

second is the intensity of light.

entering the pupil per second is given as:

s the one found in problem (a), but it is large endividual photons.

th 2271 Å from a 100 W mercury source irradiaetal. If the stopping potential is −1.3 V, estimat

ould the photo-cell respond to a high intensityÅ produced by a He-Ne laser?

upil.

ough for the

tes a photo- the work

∼105 W m−2)



Wavelength of ultraviolet light, λ = 2271 Å = 2271 × 10−10 m

Stopping potential of the metal, V 0 = 1.3 V

Planck’s constant, h = 6.6 × 10−34

J

Charge on an electron, e = 1.6 × 10−19 C

Work function of the metal =

Frequency of light = ν

We have the photo-energy relation from the photoelectric effect as:

= hν − eV 0

Let ν0 be the threshold frequency of the metal.

∴ = hν0

Wavelength of red light, = 6328 × 10−10 m

∴Frequency of red light,



Since ν0> νr , the photocell wi

Question 11.27:

Monochromatic radiation ofirradiates photosensitive matmeasured to be 0.54 V. Theirradiates the same photo-cel

Answer

Wavelength of the monochr

= 640.2 × 10−9 m

Stopping potential of the neo



Let be the work function

We have the photo-energy re

eV 0 = hν −

Wavelength of the radiation

ll not respond to the red light produced by the l

avelength 640.2 nm (1nm = 10−9

m) from a neerial made of caesium on tungsten. The stoppinource is replaced by an iron source and its 427.l. Predict the new stopping voltage.

matic radiation, λ = 640.2 nm

n lamp, V 0 = 0.54 V

6 × 10−19 C

0−34 Js

nd ν be the frequency of emitted light.

lation from the photoelectric effect as:

mitted from an iron source, λ' = 427.2 nm

ser.

on lampvoltage isnm line



= 427.2 × 10−9 m

Let be the new stopping p

Hence, the new stopping pot

Question 11.28:

A mercury lamp is a conveni photoelectric emission, sincethe red end of the visible spefollowing lines from a mercu

λ1 = 3650 Å, λ2= 4047 Å, λ3

The stopping voltages, respe

V 01 = 1.28 V, V 02 = 0.95 V,

Determine the value of Plancfor the material.

[ Note: You will notice that tcan take to be 1.6 × 10−19 C).

by Millikan, who, using hisEinstein’s photoelectric equathe value of h.]

Answer

otential. Hence, photo-energy is given as:

ntial is 1.50 eV.

ent source for studying frequency dependence oit gives a number of spectral lines ranging fro

ctrum. In our experiment with rubidium photo-cry source were used:

4358 Å, λ4= 5461 Å, λ5= 6907 Å,

tively, were measured to be:

03 = 0.74 V, V 04 = 0.16 V, V 05 = 0 V

k’s constant h, the threshold frequency and wor

get h from the data, you will need to know e (Experiments of this kind on Na, Li, K, etc. werwn value of e (from the oil-drop experiment) ction and at the same time gave an independent

fthe UV to

ell, the

k function

hich youe performednfirmedstimate of



Einstein’s photoelectric equation is given as:

eV 0 = hν−

Where,

V 0 = Stopping potential


e = Charge on an electron

ν = Frequency of radiation

= Work function of a material

It can be concluded from equation (1) that potential V 0 is directly proportional tofrequency ν.

Frequency is also given by the relation:

This relation can be used to obtain the frequencies of the various lines of the givenwavelengths.

The given quantities can be listed in tabular form as:



Frequency × 1014

Hz 8.219 7.412 6.884 5.493 4.343

Stopping potential V 0 1.28 0.95 0.74 0.16 0

The following figure shows a graph between νand V 0.

It can be observed that the obtained curve is a straight line. It intersects the ν-axis at 5 ×1014 Hz, which is the threshold frequency (ν0) of the material. Point D corresponds to afrequency less than the threshold frequency. Hence, there is no photoelectric emission forthe λ5 line, and therefore, no stopping voltage is required to stop the current.

Slope of the straight line =

From equation (1), the slope can be written as:

The work function of the metal is given as:

= h ν0

= 6.573 × 10−34 × 5 × 1014

= 3.286 × 10−19 J

= 2.054 eV



Question 11.29:

The work function for the fol

Na: 2.75 eV; K: 2.30 eV; M photoelectric emission for am away from the photocell?away?

Answer

Mo and Ni will not show ph

Wavelength for a radiation,



The energy of incident radiat

It can be observed that the efunction of Na and K only. It

photoelectric emission.

If the source of light is brougthen the intensity of radiatioradiation. Hence, the resultemitted from Na and K will i

Question 11.30:

lowing metals is given:

: 4.17 eV; Ni: 5.15 eV. Which of these metalsadiation of wavelength 3300 Å from a He-Cd l

hat happens if the laser is brought nearer and

toelectric emission in both cases

= 3300 Å = 3300 × 10−10 m

s

0−34 Js

ion is given as:

ergy of the incident radiation is greater than theis less for Mo and Ni. Hence, Mo and Ni will n

ht near the photocells and placed 50 cm away fwill increase. This does not affect the energy o

ill be the same as before. However, the photoelncrease in proportion to intensity.

ill not giveser placed 1laced 50 cm

workot show

om them,f theectrons



Light of intensity 10−5 W m−2 falls on a sodium photo-cell of surface area 2 cm2.Assuming that the top 5 layers of sodium absorb the incident energy, estimate timerequired for photoelectric emission in the wave-picture of radiation. The work functionfor the metal is given to be about 2 eV. What is the implication of your answer?

Answer

Intensity of incident light, I = 10−5 W m−2

Surface area of a sodium photocell, A = 2 cm2 = 2 × 10−4 m2

Incident power of the light, P = I × A

= 10−5 × 2 × 10−4

= 2 × 10−9 W

Work function of the metal, = 2 eV

= 2 × 1.6 × 10−19

= 3.2 × 10−19 J

Number of layers of sodium that absorbs the incident energy, n = 5

We know that the effective atomic area of a sodium atom, Ae is 10−20 m2.

Hence, the number of conduction electrons in n layers is given as:

The incident power is uniformly absorbed by all the electrons continuously. Hence, theamount of energy absorbed per second per electron is:



Time required for photoelect

The time required for the ph practical. Hence, the wave pi

Question 11.31:

Crystal diffraction experimethrough appropriate voltage.comparison, take the waveleatomic spacing in the lattice)

Answer

An X-ray probe has a greate

Wavelength of light emitted

Mass of an electron, me = 9.1



The kinetic energy of the ele

Where,

v = Velocity of the electron

ic emission:

toelectric emission is nearly half a year, whichcture is in disagreement with the given experim

ts can be performed using X-rays, or electronsWhich probe has greater energy? (For quantitatgth of the probe equal to 1 Å, which is of the o(me= 9.11 × 10−31 kg).

energy than an electron probe for the same wa

from the probe, λ = 1 Å = 10−10 m

1 × 10−31 kg

0−34 Js

6 × 10−19 C

tron is given as:

is notent.

acceleratedveder of inter-

elength.



mev = Momentum ( p) of the

According to the de Broglie

Energy of a photon,

Hence, a photon has a greate

Question 11.32:

Obtain the de Broglie wavelseen in Exercise 11.31, an elexperiments. Would a neutro(mn= 1.675 × 10−27 kg)

Obtain the de Broglie wavel(27 ºC). Hence explain whyenvironment before it can be

Answer

De Broglie wavelength =experiment

lectron

rinciple, the de Broglie wavelength is given as:

energy than an electron for the same waveleng

ngth of a neutron of kinetic energy 150 eV. Asctron beam of this energy is suitable for crystal

n beam of the same energy be equally suitable?

ngth associated with thermal neutrons at roomfast neutron beam needs to be thermalised wit

used for neutron diffraction experiments.

; neutron is not suitable for the dif

th.

you havediffractionExplain.

emperaturethe

fraction



Kinetic energy of the neutron, K = 150 eV

= 150 × 1.6 × 10−19

= 2.4 × 10−17 J

Mass of a neutron, mn = 1.675 × 10−27

kg

The kinetic energy of the neutron is given by the relation:

Where,

v = Velocity of the neutron

mnv = Momentum of the neutron

De-Broglie wavelength of the neutron is given as:

It is given in the previous problem that the inter-atomic spacing of a crystal is about 1 Å,i.e., 10−10 m. Hence, the inter-atomic spacing is about a hundred times greater. Hence, aneutron beam of energy150 eV is not suitable for diffraction experiments.

De Broglie wavelength =

Room temperature, T = 27°C = 27 + 273 = 300 K

The average kinetic energy of the neutron is given as:

Where,



k = Boltzmann constant = 1.

The wavelength of the neutr

This wavelength is comparabenergy neutron beam should

Question 11.33:

An electron microscope usesde Broglie wavelength assocaperture, etc.) are taken to beelectron microscope compar

Answer

Electrons are accelerated by


Mass of an electron, me = 9.1

Wavelength of yellow light

The kinetic energy of the ele

E = eV

= 1.6 × 10−19 × 50 × 103

= 8 × 10−15 J

De Broglie wavelength is giv

8 × 10−23 J mol−1 K −1

n is given as:

le to the inter-atomic spacing of a crystal. Hencfirst be thermalised, before using it for diffracti

electrons accelerated by a voltage of 50 kV. Deated with the electrons. If other factors (such asroughly the same, how does the resolving powwith that of an optical microscope which uses

a voltage, V = 50 kV = 50 × 103 V

6 × 10−19 C

1 × 10−31 kg

5.9 × 10−7 m

tron is given as:

en by the relation:

e, the high- n.

termine thenumerical

r of anellow light?



This wavelength is nearly 10

The resolving power of a miused. Thus, the resolving pooptical microscope.

Question 11.34:

The wavelength of a probe isin some detail. The quark strscale of 10−15 m or less. Thiselectron beams produced byhave been the order of energ0.511 MeV.)

Answer

Wavelength of a proton or a

Rest mass energy of an elect

m0c2 = 0.511 MeV

= 0.511 × 106 × 1.6 × 10−19

= 0.8176 × 10−13

J



The momentum of a proton

times less than the wavelength of yellow light.

roscope is inversely proportional to the waveleer of an electron microscope is nearly 105 time

roughly a measure of the size of a structure thacture of protons and neutrons appears at the mistructure was first probed in early 1970’s using

a linear accelerator at Stanford, USA. Guess whof these electron beams. (Rest mass energy of

eutron, λ ≈ 10−15 m

on:

0−34 Js

s

r a neutron is given as:

gth of lights that of an

t it can probenute length-

high energyat mightelectron =



The relativistic relation for e

Thus, the electron energy emorder of 1.24 BeV.

Question 11.35:

Find the typical de Broglie wtemperature (27 ºC) and 1 attwo atoms under these condi

Answer

De Broglie wavelength assoc

Room temperature, T = 27°C

Atmospheric pressure, P = 1

ergy ( E ) is given as:

itted from the accelerator at Stanford, USA mig

avelength associated with a He atom in helium pressure; and compare it with the mean separ

ions.

iated with He atom =

= 27 + 273 = 300 K

atm = 1.01 × 105 Pa

ht be of the

gas at roomtion between



Atomic weight of a He atom = 4

Avogadro’s number, NA = 6.023 × 1023

Boltzmann constant, k = 1.38 × 10−23 J mol−1 K −1

Average energy of a gas at temperature T ,is given as:

De Broglie wavelength is given by the relation:

Where,

m = Mass of a He atom

We have the ideal gas formula:

PV = RT

PV = kNT

Where,

V = Volume of the gas

N = Number of moles of the gas



Mean separation between tw

Hence, the mean separationwavelength.

Question 11.36:

Compute the typical de Brogcompare it with the mean se be about 2 × 10−10 m.

[ Note: Exercises 11.35 and 1gaseous molecules under ord

packets in a metal strongly omolecules in an ordinary gasdistinguished apart from oneimplications which you will

Answer

Temperature, T = 27°C = 27

Mean separation between tw

De Broglie wavelength of an

Where,

h = Planck’s constant = 6.6 ×

atoms of the gas is given by the relation:

etween the atoms is much greater than the de B

lie wavelength of an electron in a metal at 27 ºCaration between two electrons in a metal which

1.36 reveal that while the wave-packets associainary conditions are non-overlapping, the electrerlap with one another. This suggests that whecan be distinguished apart, electrons in a metalanother. This indistinguishibility has many fun

xplore in more advanced Physics courses.]

+ 273 = 300 K

electrons, r = 2 × 10−10 m

electron is given as:

10−34 Js

roglie

andis given to

ed withon wave-

eascannot beamental



m = Mass of an electron = 9.

k = Boltzmann constant = 1.

Hence, the de Broglie wavelseparation.

Question 11.37:

Answer the following questi

Quarks inside protons and ne(−1/3)e]. Why do they not sh

What is so special about theseparately?

Why should gases be insulat pressures?

Every metal has a definite wthe same energy if incident r

distribution of photoelectron

The energy and momentumthe associated matter wave b

E = hν, p =

But while the value of λ is pof the phase speed νλ) has no

Answer

11 × 10−31 kg

8 × 10−23 J mol−1 K −1

ngth is much greater than the given inter-electr

ns:

utrons are thought to carry fractional charges [( ow up in Millikan’s oil-drop experiment?

ombination e/m? Why do we not simply talk o

rs at ordinary pressures and start conducting at

rk function. Why do all photoelectrons not codiation is monochromatic? Why is there an ene

?

f an electron are related to the frequency and wthe relations:

ysically significant, the value of ν (and therefor physical significance. Why?

on

2/3)e ;

e and m

very low

e out withrgy

avelength of

e, the value



Quarks inside protons and neutrons carry fractional charges. This is because nuclear forceincreases extremely if they are pulled apart. Therefore, fractional charges may exist innature; observable charges are still the integral multiple of an electrical charge.

The basic relations for electric field and magnetic field are

.

These relations include e (electric charge), v (velocity), m (mass), V (potential), r (radius),and B (magnetic field). These relations give the value of velocity of an electron as

and

It can be observed from these relations that the dynamics of an electron is determined not by e and m separately, but by the ratio e/m.

At atmospheric pressure, the ions of gases have no chance of reaching their respectiveelectrons because of collision and recombination with other gas molecules. Hence, gasesare insulators at atmospheric pressure. At low pressures, ions have a chance of reachingtheir respective electrodes and constitute a current. Hence, they conduct electricity atthese pressures.

The work function of a metal is the minimum energy required for a conduction electron to

get out of the metal surface. All the electrons in an atom do not have the same energylevel. When a ray having some photon energy is incident on a metal surface, the electronscome out from different levels with different energies. Hence, these emitted electronsshow different energy distributions.

The absolute value of energy of a particle is arbitrary within the additive constant. Hence,wavelength ( λ) is significant, but the frequency (ν) associated with an electron has nodirect physical significance.

Therefore, the product νλ(phase speed)has no physical significance.

Group speed is given as:



This quantity has a physical eaning.

chapter 11 dual nature of radiation and matter

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